Answer:
The correct answer would be - 0.1 a, 0.9 A
Step-by-step explanation:
The initial population of the colonist had free earlobes which are represented by A here as it is dominant and attached is represent by a. purebred dominant would have AA.
Frequency of allele A =
8 pure dominant and 2 heterozygous dominant presents in the population
So, frequency of A allele = 8 x 2 + 2 / 20
= 18/20
= 0.9
As we know
p+q=1;
frequency of allele "a" = 1-p
= 1-0.9
= 0.1